hmb
- 15
- 0
Homework Statement
Prove:
Let F and G be functions. F = G if and only if domF = domG and F(x) = G(x) for all x \in domF.
Homework Equations
?
The Attempt at a Solution
If F = G, then (xFy if and only if xGy) (Substitutivity of identicals)
If (xFy if and only if xGy), then F = G (Axiom of extensionality)
So F = G if and only if (xFy if and only if xGy)
Since F and G are functions,
For all x \in domF, xFy if and only if y = F(x)
For all x \in domG, xGy if and only if y = G(x)
xFy if and only if x \in domF and xFy
xGy if and only if x \in domG and xGy
xFy if and only if x \in domF and y = F(x)
xGy if and only if x \in domG and y = G(x)
So
F = G if and only if ((x \in domF and y = F(x)) if and only if (x \in domG and y = G(x)))
which is equivalent to
F = G if and only if ((x \in domF and x \in domG and y = F(x) and y = G(x)) or (x \in domF and x \in domG and y \neq F(x) and y \neq G(x)) or (x \notin domF and x \notin domG) or (x \in domF and x \notin domG and y \neq F(x)) or (x \notin domF and x \in domG and y \neq G(x))
I was hoping I could show that domF = domG and F(x) = G(x) for all x \in domF was equivalent to this but I couldn't. This is as close as I could get:
(domF = domG and F(x) = G(x) for all x \in domF) if and only if (x \in domF if and only if x \in domG and, if y \in domF, then y = F(x) if and only if y = G(x))
(domF = domG and F(x) = G(x) for all x \in domF) if and only if (((x \in domF and x \in domG and y = F(x) and y = G(x)) or (x \in domF and x \in domG and y \neq F(x) and y \neq G(x)) or (x \notin domF and x \notin domG))
This is close to what I had above but not quite.
I don't know. I think I need to get a set theory book that has solutions; I am currently working through "Introduction to Set Theory" by Hrbacek and Jech. I might switch to "Naive Set Theory" by Paul Hamos; it doesn't even have exercises I don't think.
Any help would be appreciated.